f(x)=sinx-√3cosx=2[(1/2)sinx-(√3/2)cosx]
=2[sinxcos(π/3)-cosxsin(π/3)]
=2sin(x-π/3)
令-π/2+2kπ≤x-π/3≤π/2+2kπ.k∈Z
即-π/6+2kπ≤x≤5π/6+2kπ.k∈Z
又因为x∈[-π,0]
令k=0可得递增区间为[-π/6,0]